Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, p\neq 0$. $\dfrac{{(k^{-3})^{4}}}{{(k^{-4}p^{-2})^{-1}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{-3}}$ to the exponent ${4}$ . Now ${-3 \times 4 = -12}$ , so ${(k^{-3})^{4} = k^{-12}}$ In the denominator, we can use the distributive property of exponents. ${(k^{-4}p^{-2})^{-1} = (k^{-4})^{-1}(p^{-2})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{-3})^{4}}}{{(k^{-4}p^{-2})^{-1}}} = \dfrac{{k^{-12}}}{{k^{4}p^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-12}}}{{k^{4}p^{2}}} = \dfrac{{k^{-12}}}{{k^{4}}} \cdot \dfrac{{1}}{{p^{2}}} = k^{{-12} - {4}} \cdot p^{- {2}} = k^{-16}p^{-2}$.